Question Number 169501 by Giantyusuf last updated on 01/May/22
Answered by qaz last updated on 01/May/22
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\:\mathrm{n}\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}narctan}\:\frac{\mathrm{1}}{\mathrm{n}}=\mathrm{1} \\ $$